[math-fun] iterating the inverse of a spacefill converges quadratically near its fixed pt.

E.g., the leading octal digits of the middle fixed point of the Dragon: In[36]:= FromDigits[#, 8]/8^Length[#] &@{6, 4, 1, 7, 6, 4, 0, 0, 0, 3, 2, 0, 0, 0, 0, 0, 0, 0, 1, 5, 0} Out[36]= 941199549051961357/1152921504606846976 The inverse In[37]:= undrag[%] Out[37]= {1085129200228273162089124938402233613/ 1329227995784915872903807060280344576} is unique. In[38]:= RealDigits[%[[1]], 8] Out[38]= {{6, 4, 1, 7, 6, 4, 0, 0, 0, 3, 2, 0, 0, 0, 0, 0, 0, 0, 1, 5, 0, 3, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 4, 1, 5}, 0} And *all* these digits are good! Another e.g.: The leading hex digits of Thue-Morse (HAKMEM's "parity constant"): (1/2)*(2 - Product[1 - 2^(-2^k), {k, 0, Infinity}]) In[31]:= %/2 /.∞-> 6 Out[31]= 280701669626288379716180548005699333331/ 680564733841876926926749214863536422912 In[32]:= RealDigits[%, 16] Out[32]= {{6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 8}, 0} The exact constant satisfies hilbert[x]==I+x . Inverting I+x, In[33]:= unbert[%% + I] Out[33]= { 573106971234556259792196936825953480449973086345124920770960412939461034308365/ 1389505070847794345082851820104254894239239815987686768473491008094957555679232, 573106971234556259792196936825953480449973086345124920770960412939461034308367/ 1389505070847794345082851820104254894239239815987686768473491008094957555679232} Two very close values In[34]:= RealDigits[%[[1]], 16] Out[34]= {{6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 6, {10}}, 0} In[39]:= Length[Out[#][[1]]] & /@ {32, 34} Out[39]= {33, 66} Digits doubled. Intuitively, you need 2n bits of t to get n bits of each of the imaginary and real parts of spacefill(t). When we take the inverse of x+0i or x+I, we are actually supplying n bits of x and n bits of 0 or 1, so we've cranked in 2n bits, and we get them back. --rwg Note that 1-Thue also satisfies hilbert[x]==I+x due to the left-right symmetry of the function: hilbert[1-t] = 1-Conjugate[hilbert[t]] . "That can't be right." But it is.